C到MIPS汇编语言

Question

i and j are assigned to registers $s3 and $s4 and the base address of A and B are in registers $s6 and $s7 .

B [8] = A [i-j]

So the answer is something like

sub $t0, $s3, $s4
add $t0, $s6, $t0
lw $t1, 16($t0)   (What is happening here.. I am so confused)
sw $t1, 32($s7)

Please explain. I am so confused

Answer 1

My MIPS assembly is rusty, but...

lw/sw are load word/store word.

syntax being lw register, address

So for the instruction you indicated, the address in $t0 is being loaded into $t1. The 16 thing is a notation to add 16 to $t0 before the load word instruction.

Answer 2

The code looks incorrect:

sub $t0, $s3, $s4  # t0 = i -j
# there should be a "sll $t0, 2" here to convert i-j to a byte offset
add $t0, $s6, $t0  # t0 += &A
lw $t1, 16($t0)    # this line is incorrect - it should be lw t1,($t0) with no offset
sw $t1, 32($s7)    # store t1 at &B + 8 words i.e. 32 bytes